用回朔的办法解递归 L(m)=L(-1)+n =L(n-2)+(n-1)+n =L(n-3)+(m-2)+(n-1)+n =L(0)+1+2+..+(n-2)t(n-1)tn L(m=n(n+)/2+1
用回朔的办法解递归 L(n) = L(n-1)+n = L(n-2)+(n-1)+n = L(n-3)+(n-2)+(n-1)+n = …… = L(0)+1+2+……+(n-2)+(n-1)+n L(n) = n(n+1)/2 + 1
递归思维：Josephus问题 Live or die,it's a problem! Legend has it that Josephus wouldn't have lived to become famous without his mathematical talents. During the Jewish Roman war,he was among a band of 41 Jewish rebels trapped in a cave by the Romans. Preferring suicide to capture,the rebels decided to form a circle and, proceeding around it,to killevery third emaining person until no one was left. 0 But Josephus,along with an unindicted co-conspirator,wanted none of this suicide nonsense;so he quickly calcdlated where he and his friend should stand in the vicious circle. We use a simpler version: “every second
递归思维：Josephus 问题 ◼ Live or die, it’s a problem! ❑ Legend has it that Josephus wouldn't have lived to become famous without his mathematical talents. ❑ During the Jewish Roman war, he was among a band of 41 Jewish rebels trapped in a cave by the Romans. ❑ Preferring suicide to capture, the rebels decided to form a circle and, proceeding around it, to kill every third remaining person until no one was left. ❑ But Josephus, along with an unindicted co-conspirator, wanted none of this suicide nonsense; so he quickly calculated where he and his friend should stand in the vicious circle. We use a simpler version: “every second
试试看：n=10 一及 survivor J(10)=5
试试看：n=10 1 8 7 6 2 4 5 10 3 9 1 7 5 9 3 1 7 5 9 3 survivor J(10)=5